Mixing
A Topological Invariant for $pi_3(U(n))$
$begingroup$
Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$:
$$
N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] ,
$$
where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$, $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$, and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why?
Update 02/02/2019
I think I got an argument for $n=2$. In this case, $U=e^{i varphi} q$ with $qin SU(2)$.
Due to the trace and the Levi-Civita symbol in $N$, $varphi$ does not contribute to $N$. As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$, $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partial_i q=i(A_i sigma_x +B_i sigma_y+C_i sigma_z )$ with $A_i,B_i,C_i$ real and $sigma$'s being the Pauli matrices. As a result, we have
$$
N=frac{1}{2pi^2}int_{S^3} d^3x epsilon^{ijk} A_i B_i C_i .
$$
Furthermore, $SU(2)$ is diffeomorphic to $S^3$, and thus $q$ can be parametrized as the three angles $boldsymbol{Omega}=(psi,theta,phi)$ of $S^3$. By choosing the right convention, $N$ can be further written as
$$
N=frac{1}{2pi^2}int_{S^3} d^3x left|frac{partialboldsymbol{Omega}}{partialboldsymbol{x}}right| sin(theta)sin^2(phi) ,
$$
where $left|frac{partialboldsymbol{Omega}}{partialboldsymbol{x}}right|$ is the determinant of the Jacobian matrix.
Clearly, $N$ indicates how many times the $boldsymbol{x}$-$S^3$ warps around the $boldsymbol{Omega}$-$S^3$.
For $n>2$, I may guess that some $SU(2)$ subgroup of $U(n)$ accounts for $N$, but I can be wrong.
multivariable-calculus algebraic-topology differential-topology lie-groups vector-bundles
asked Feb 1 at 3:54
KarlKarl
162
$endgroup$
add a comment |
$begingroup$
Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$:
$$
N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] ,
$$
where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$, $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$, and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why?
Update 02/02/2019
I think I got an argument for $n=2$. In this case, $U=e^{i varphi} q$ with $qin SU(2)$.
Due to the trace and the Levi-Civita symbol in $N$, $varphi$ does not contribute to $N$. As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$, $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partial_i q=i(A_i sigma_x +B_i sigma_y+C_i sigma_z )$ with $A_i,B_i,C_i$ real and $sigma$'s being the Pauli matrices. As a result, we have
$$
N=frac{1}{2pi^2}int_{S^3} d^3x epsilon^{ijk} A_i B_i C_i .
$$
Furthermore, $SU(2)$ is diffeomorphic to $S^3$, and thus $q$ can be parametrized as the three angles $boldsymbol{Omega}=(psi,theta,phi)$ of $S^3$. By choosing the right convention, $N$ can be further written as
$$
N=frac{1}{2pi^2}int_{S^3} d^3x left|frac{partialboldsymbol{Omega}}{partialboldsymbol{x}}right| sin(theta)sin^2(phi) ,
$$
where $left|frac{partialboldsymbol{Omega}}{partialboldsymbol{x}}right|$ is the determinant of the Jacobian matrix.
Clearly, $N$ indicates how many times the $boldsymbol{x}$-$S^3$ warps around the $boldsymbol{Omega}$-$S^3$.
For $n>2$, I may guess that some $SU(2)$ subgroup of $U(n)$ accounts for $N$, but I can be wrong.
multivariable-calculus algebraic-topology differential-topology lie-groups vector-bundles
asked Feb 1 at 3:54
KarlKarl
162
$endgroup$
add a comment |
$begingroup$
Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$:
$$
N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] ,
$$
where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$, $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$, and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why?
Update 02/02/2019
I think I got an argument for $n=2$. In this case, $U=e^{i varphi} q$ with $qin SU(2)$.
Due to the trace and the Levi-Civita symbol in $N$, $varphi$ does not contribute to $N$. As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$, $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partial_i q=i(A_i sigma_x +B_i sigma_y+C_i sigma_z )$ with $A_i,B_i,C_i$ real and $sigma$'s being the Pauli matrices. As a result, we have
$$
N=frac{1}{2pi^2}int_{S^3} d^3x epsilon^{ijk} A_i B_i C_i .
$$
Furthermore, $SU(2)$ is diffeomorphic to $S^3$, and thus $q$ can be parametrized as the three angles $boldsymbol{Omega}=(psi,theta,phi)$ of $S^3$. By choosing the right convention, $N$ can be further written as
$$
N=frac{1}{2pi^2}int_{S^3} d^3x left|frac{partialboldsymbol{Omega}}{partialboldsymbol{x}}right| sin(theta)sin^2(phi) ,
$$
where $left|frac{partialboldsymbol{Omega}}{partialboldsymbol{x}}right|$ is the determinant of the Jacobian matrix.
Clearly, $N$ indicates how many times the $boldsymbol{x}$-$S^3$ warps around the $boldsymbol{Omega}$-$S^3$.
For $n>2$, I may guess that some $SU(2)$ subgroup of $U(n)$ accounts for $N$, but I can be wrong.
multivariable-calculus algebraic-topology differential-topology lie-groups vector-bundles
asked Feb 1 at 3:54
KarlKarl
162
$endgroup$
Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$:
$$
N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] ,
$$
where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$, $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$, and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why?
Update 02/02/2019
I think I got an argument for $n=2$. In this case, $U=e^{i varphi} q$ with $qin SU(2)$.
Due to the trace and the Levi-Civita symbol in $N$, $varphi$ does not contribute to $N$. As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$, $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partial_i q=i(A_i sigma_x +B_i sigma_y+C_i sigma_z )$ with $A_i,B_i,C_i$ real and $sigma$'s being the Pauli matrices. As a result, we have
$$
N=frac{1}{2pi^2}int_{S^3} d^3x epsilon^{ijk} A_i B_i C_i .
$$
Furthermore, $SU(2)$ is diffeomorphic to $S^3$, and thus $q$ can be parametrized as the three angles $boldsymbol{Omega}=(psi,theta,phi)$ of $S^3$. By choosing the right convention, $N$ can be further written as
$$
N=frac{1}{2pi^2}int_{S^3} d^3x left|frac{partialboldsymbol{Omega}}{partialboldsymbol{x}}right| sin(theta)sin^2(phi) ,
$$
where $left|frac{partialboldsymbol{Omega}}{partialboldsymbol{x}}right|$ is the determinant of the Jacobian matrix.
Clearly, $N$ indicates how many times the $boldsymbol{x}$-$S^3$ warps around the $boldsymbol{Omega}$-$S^3$.
For $n>2$, I may guess that some $SU(2)$ subgroup of $U(n)$ accounts for $N$, but I can be wrong.
multivariable-calculus algebraic-topology differential-topology lie-groups vector-bundles
multivariable-calculus algebraic-topology differential-topology lie-groups vector-bundles
asked Feb 1 at 3:54
KarlKarl
162
asked Feb 1 at 3:54
KarlKarl
162
edited Feb 3 at 3:16
Karl
asked Feb 1 at 3:54
KarlKarl
162
asked Feb 1 at 3:54
KarlKarl
162
asked Feb 1 at 3:54
KarlKarl
162
162
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